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2. Fie triunghiul dreptunghic ABC, m(Â)=90°, AB=3 cm, AC= 3√3 cm. Determinați lungimea laturii BC precum şi: sin (<B), cos(<B), tg (<B),ctg(<B) ​

Răspuns :

Funcțiile trigonometrice:

[tex]\boxed{\text{sin}=\dfrac{\text{cateta opus}\breve{\text{a}}}{\text{ipotenuz}\breve{\text{a}}}} \ , \ \boxed{\text{cos}=\dfrac{\text{cateta al}\breve{\text{a}}\text{turat}\breve{\text{a}}}{\text{ipotenuz}\breve{\text{a}}}} \\ \boxed{\text{tg}=\dfrac{\text{cateta opus}\breve{\text{a}}}{\text{cateta al}\breve{\text{a}}\text{turat}\breve{\text{a}}}} \ , \boxed{ \text{ctg}=\dfrac{\text{cateta al}\breve{\text{a}}\text{turat}\breve{\text{a}}}{\text{cateta opus}\breve{\text{a}}}}[/tex]

Avem ΔABC dreptunghic, AB=3 cm, AC=[tex]3\sqrt{3}[/tex] cm. Pentru a afla BC aplicăm Teorema lui Pitagora:

[tex]BC^2=AB^2+AC^2\\ BC^2=3^2+(3\sqrt{3})^2\\ BC^2=9+9\cdot 3\\ BC^2=9+27\\ BC^2=36\\ \tt \Rightarrow BC=6 \ cm[/tex]

Aplicăm funcțiile trigonometrice pentru ∡B.

[tex]\sin \measuredangle B=\dfrac{AC}{BC}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\\ \cos \measuredangle B=\dfrac{AB}{BC}=\dfrac{3}{6}=\dfrac{1}{2}\\ \text{tg} \measuredangle B=\dfrac{AC}{AB}=\dfrac{3\sqrt{3}}{3}=\sqrt{3}\\ \text{ctg}\measuredangle B=\dfrac{AB}{AC}=\dfrac{3}{3\sqrt{3}}=\dfrac{\sqrt{3}}{3}[/tex]

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